Find the equation of the normal to the ellipse `(x^2)/(a^2)+(y^2)/(b^2)=1` at the positive end of the latus rectum.

Options

A. `x+ey+e^(3)a=0`
B. `x-ey-e^(3)a=0`
C. `x-ey-e^(2)a=0`
D. none of these

Answer

B

Explanation

the equation of the normal at `(x_(1),y_(1))` to the given ellipse is
`(a^(2)x)/(x_(1))-(b^(2)y)/(y_(1))=a^(2)-b^(2)`
Here,`x_(1)=ae and y_(1)=(b^(2))/(a).`
SO , the equation of the normal at positive end of the latusrectum is
`(a^(2)x)/(ae)-(b^(2)y)/(b^(2)//a)=a^(2)e^(2) [:' b^(2)=a^(2)(1-e^(2))]`
`implies (ax)/(e) -ay=a^(2)e^(2)implies x=ey -e^(3)a=0`